Theorem.
AP BQ CR


4
AD BE CF
Proof. The area of ABD, denoted by A, is

A = 12 BCAD = 12 AC BE = 12 ABCF
because AD, CF, and BE are altitudes. Therefore

2A = BCAD= AC BE = ABCF


and

AD =
2A
BC
BE =
2A
AC
CF =
2A
AB
By substituting AD, BE, and CF by
the fractions as above, the expression
AP BQ CR
is equivalent to 


AD BE CF

AP BQ CR AP  BC  BQ  AC  CR AB



AD BE CF
2A
To prove the theorem, we need to show that APBC + BQAC + CRAB = 8A.
Let’s observe APBC with respect to the area of ABD. Four parallel lines to AP and

BC respectively form a rectangle abde and its area is APBC. Moreover,
AP  BC  2A  DP  BC

Similarly, we have
BQAC = 2A + EQAC
CRAB = 2A + FRAB
Therefore we need to show that
DPBC + EQAC + FRAB = 2A
because the expression
AP BQ CR AP  BC  BQ  AC  CR AB



AD BE CF
2A

6A  DP  BC  EQ AC  FR AB
2A
should be 
4. To do so, we need the following lemma.

Lemma. DP = HD, EQ = HE, and FR = HF.
Proof. We will show EQ = HE by using congruency of two triangles CHE and CQE.
The other two equalities can be proved in a similar manner.
The angles BAC and BQC are the same, say , because they share the same arc on
the circumcised circle. Since HFA and HEA are the right angles, FAE and FHE
are supplementary (they add up to 180 degrees). From the figure above, we see
that EHC is also a supplementary angle of FHE, and hence EHC = . Since
EHC =  = EQC, two triangles CHE and CQE are congruent (ASA). Then EQ and
HE are equal as the corresponding sides of two congruent triangles. QED.
Let us go back the theorem. Our claim is: DPBC + EQAC + FRAB = 2A. We note that
ABC consists of three triangles: HBC, AHC, and AHB.
By Lemma above, we have
DPBC + EQAC + FRAB = HDBC + EHAC + FHAB
=2( 12 HDBC +
1
2
EHAC +
We are done with the proof. QED.



1
2
FHAB) = 2area of ABC = 2A.